Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

double p[3005],dp[3005][3005];
int main()
{
	int n,odd;
	scanf("%d",&n);
	for(int i=0;i<n;++i)
	scanf("%lf",p+i);
	odd = n/2 + 1;
	for(int i=n;i>=0;--i)
	{	
		for(int j=0;j<=n;++j)
		{
			if(odd <= i && j == 0)
			dp[i][j] = 1;
			else if(odd > i && j == 0)
			dp[i][j] = 0;
			else
			dp[i][j] = p[j-1]*dp[i+1][j-1] + (1-p[j-1])*dp[i][j-1];
		}
	}
	printf("%0.10lf\n",dp[0][n]);
	return 0;
}